3.19.34 \(\int \frac {e^{-5-\frac {x^2}{e^5}} (e^5-2 x^2+2 x^3)}{4-8 x+4 x^2} \, dx\)

Optimal. Leaf size=22 \[ \frac {e^{-\frac {x^2}{e^5}} x}{4 (1-x)} \]

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Rubi [A]  time = 0.07, antiderivative size = 33, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 12, 2288} \begin {gather*} \frac {e^{-\frac {x^2}{e^5}} \left (x^2-x^3\right )}{4 (1-x)^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-5 - x^2/E^5)*(E^5 - 2*x^2 + 2*x^3))/(4 - 8*x + 4*x^2),x]

[Out]

(x^2 - x^3)/(4*E^(x^2/E^5)*(1 - x)^2*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-5-\frac {x^2}{e^5}} \left (e^5-2 x^2+2 x^3\right )}{4 (-1+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{-5-\frac {x^2}{e^5}} \left (e^5-2 x^2+2 x^3\right )}{(-1+x)^2} \, dx\\ &=\frac {e^{-\frac {x^2}{e^5}} \left (x^2-x^3\right )}{4 (1-x)^2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.91 \begin {gather*} -\frac {e^{-\frac {x^2}{e^5}} x}{4 (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-5 - x^2/E^5)*(E^5 - 2*x^2 + 2*x^3))/(4 - 8*x + 4*x^2),x]

[Out]

-1/4*x/(E^(x^2/E^5)*(-1 + x))

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fricas [A]  time = 0.86, size = 23, normalized size = 1.05 \begin {gather*} -\frac {x e^{\left (-{\left (x^{2} + 5 \, e^{5}\right )} e^{\left (-5\right )} + 5\right )}}{4 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)+2*x^3-2*x^2)/(4*x^2-8*x+4)/exp(5)/exp(x^2/exp(5)),x, algorithm="fricas")

[Out]

-1/4*x*e^(-(x^2 + 5*e^5)*e^(-5) + 5)/(x - 1)

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giac [A]  time = 0.23, size = 16, normalized size = 0.73 \begin {gather*} -\frac {x e^{\left (-x^{2} e^{\left (-5\right )}\right )}}{4 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)+2*x^3-2*x^2)/(4*x^2-8*x+4)/exp(5)/exp(x^2/exp(5)),x, algorithm="giac")

[Out]

-1/4*x*e^(-x^2*e^(-5))/(x - 1)

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maple [A]  time = 0.38, size = 17, normalized size = 0.77

method	result	size
risch	\(-\frac {x \,{\mathrm e}^{-x^{2} {\mathrm e}^{-5}}}{4 \left (x -1\right )}\)	\(17\)
gosper	\(-\frac {x \,{\mathrm e}^{-x^{2} {\mathrm e}^{-5}}}{4 \left (x -1\right )}\)	\(20\)
norman	\(-\frac {x \,{\mathrm e}^{-x^{2} {\mathrm e}^{-5}}}{4 \left (x -1\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)+2*x^3-2*x^2)/(4*x^2-8*x+4)/exp(5)/exp(x^2/exp(5)),x,method=_RETURNVERBOSE)

[Out]

-1/4*x*exp(-x^2*exp(-5))/(x-1)

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maxima [A]  time = 0.62, size = 16, normalized size = 0.73 \begin {gather*} -\frac {x e^{\left (-x^{2} e^{\left (-5\right )}\right )}}{4 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)+2*x^3-2*x^2)/(4*x^2-8*x+4)/exp(5)/exp(x^2/exp(5)),x, algorithm="maxima")

[Out]

-1/4*x*e^(-x^2*e^(-5))/(x - 1)

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mupad [B]  time = 0.22, size = 18, normalized size = 0.82 \begin {gather*} -\frac {x\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{-5}}}{4\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x^2*exp(-5))*exp(-5)*(exp(5) - 2*x^2 + 2*x^3))/(4*x^2 - 8*x + 4),x)

[Out]

-(x*exp(-x^2*exp(-5)))/(4*(x - 1))

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sympy [A]  time = 0.14, size = 15, normalized size = 0.68 \begin {gather*} - \frac {x e^{- \frac {x^{2}}{e^{5}}}}{4 x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)+2*x**3-2*x**2)/(4*x**2-8*x+4)/exp(5)/exp(x**2/exp(5)),x)

[Out]

-x*exp(-x**2*exp(-5))/(4*x - 4)

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